CHM 123
Chapter 11 – Properties of Solution
Homogeneous mixtures can be classified according to the size of their particles as either
solution or colloids
Solution is the common class of homogenous mixture. They are transparent, may be
colored and can’t be physically separated.
Colloids are often murky or opaque to light. (E.g: milk or fog)
Suspensions have larger particles than colloids. Their particles separate out on standing
and are visible
11.2 – Energy changes and the solution process
A good rule of thumb, “like dissolves like”
Solution will form when the three types of intermolecular interactions are similar in kind and
magnitude
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SolutetoSolvent
>
SolutetoSolvent
=
SolutetoSolvent
<
Solute-to-Solute +
Solvent-to-Solvent
Solute-to-Solute +
Solvent-to-Solvent
Solute-to-Solute +
Solvent-to-Solvent
Solution
Forms
Solution
Forms
Solution
May or May
Not Form
Simply put, three processes affect the energetics
of the process:
_ Separation of solute particles
ΔH1( this is always endothermic)
_ Separation of solvent particles ΔH2 ( this too is
always endothermic)
_ New interactions between solute and solvent
ΔH3 ( this is always exothermic)
The overall enthalpy change associated with
these three processes :
ΔHsoln = ΔH1 + ΔH2+ ΔH3 (Hess’s Law)
The solute-solvent interactions are
greater than the sum of the
solute-solute and solvent-solvent
interactions.
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The solute-solvent interactions are less
than the sum of the solute-solute and
solvent-solvent interactions.
11.3
– Units of Concentration
Example: A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the
mass percent (% m/m) of the solution.
Example: What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C27H46O,
in 40.0 g chloroform, CHCl3?
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11.4 - Factors that affects solubility
• when there is an attraction between the particles of the solute and solvent.
• when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes
such as NaCl.
• when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or
grease.
Effect of Temperature on Solubility
Solubility curves can be used to predict whether a solution with a particular amount of solute
dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above
the line)
Solubility
• depends on temperature
• most solids increases as temperature
increases.
• Hot tea dissolves more sugar than
does cold tea because the solubility
of sugar is much greater in higher
temperature
• There is no obvious correlation between
structure and solubility or between
solubility and temperature.
• When a saturated solution is carefully
cooled, it becomes a supersaturated
solution because it contains more solute
than the solubility allows.
 Unsaturated solution: a solution containing less than the equilibrium amount of solute.
 Saturated solution: A solution containing the maximum possible amount of dissolved
solute at equilibrium.
 Supersaturated Solution: A solution containing a greater-than-equilibrium amount of solute.
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Example: The solubility of NaNO3 in water at 50oC is 110g/100g of water. In a laboratory, a
student use 230.0 g of NaNO3 with 200 g of water at the same temperature
o How many grams of NaNO3 will dissolve?
o Is the solution saturated, super saturated or unsaturated?
o What is the mass, in grams, of any solid NaNO3 on the bottom of the container?
Gases in Solution
• In general, the solubility of gases in water increases with increasing mass as the attraction between
the gas and the solvent molecule is mainly dispersion forces.
• Larger molecules have stronger dispersion forces.
Henry’s law states
• the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.
• at higher pressures, more gas molecules dissolve in the liquid.
Solubility = k ·P
where
• k is the Henry’s law constant for
that gas in that solvent at that
temperature
• P is the partial pressure of the gas
above the liquid.
Henry’s Law is what (in a manner of thinking) gives soda pop its fizz. The bubbling that occurs
when a can of soda is opened results from the reduced pressure of carbon dioxide over the liquid.
At lower pressure, the carbon dioxide is less soluble and bubbles of solution.
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Example: Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure
of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this
temperature is 3.1 x 10–2 M•atm.
11.5 – 11.10 – Physical Behavior of Solutions: Colligative Properties
Colligative Properties: Properties that depend on the amount of a dissolved solute but not on its
chemical identity.
• Vapor-Pressure Lowering
Boiling-Point Elevation
• Freezing-Point Depression
Osmotic Pressure
Solutions of ionic substances often have a vapor pressure significantly lower than predicted,
because the ion-dipole forces between the dissolved ions and polar water molecules are so strong.
Vapor-Pressure Lowering of Solutions:
Raoult’s Law
Beakers with equal liquid levels of pure
solvent and a solution are place in a bell jar.
Solvent molecules evaporate from each one
and fill the bell jar, establishing an
equilibrium with the liquids in the beakers.
When equilibrium is established, the liquid
level in the solution beaker is higher than the
solution level in the pure solvent beaker – the
thirsty solution grabs and holds solvent vapor
more effectively
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The vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal
vapor pressure, P°
◦ since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will
always be less than the vapor pressure of the pure solvent
Example: Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of
C12H22O11 with 300.0 g of H2O. PH2O = 23.8 atm
A solution consists of 3.88 g benzene,C6H6 , and 2.45 g toluene ,C6H5CH3. The vapor pressure of
pure benzene at 20.0 ° C is 75.0 mm Hg and that of toluene at 20.0 ° C is 22.0 mm Hg. Assume
that Raoult’s law holds for each component of the solution, calculate the mole fraction of benzene
in the vapor. ( molar mass of benzene= 78.0 g/mole and toluene = 92.0 g/mole.)
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Solutions with a Nonvolatile Solute
 according to Raoult’s Law, the effect of solute on the vapor pressure simply depends on the
number of solute particles
 when ionic compounds dissolve in water, they dissociate – so the number of solute particles
is a multiple of the number of moles of formula units
 the effect of ionic compounds on the vapor pressure of water is magnified by the
dissociation
◦ since NaCl dissociates into 2 ions, Na+ and Cl, one mole of NaCl lowers the vapor
pressure of water twice as much as 1 mole of C12H22O11 molecules would
When NaCl(s) dissolves in water, the reaction can
be written as
NaCl(s)
Na+(aq) + Cl-(aq)
solid
separation of ions
If an ionic salt is soluble in water, it is because the
ion dipole
Interactions are strong enough to overcome the
lattice energy of the salt crystal.
• For sodium chloride, the predicted
value of i is 2. For a 0.5 m solution
of sodium chloride, the
experimental value for i is 1.9.
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• The smaller than expected value for i
is due to incomplete dissociation of
ionic compounds. As the
concentration decreases, the
experimental value for i increases until
it reaches the theoretical maximum
(decreased occurrence of ion-pairs in
solution also plays a factor).
Example – What is the vapor pressure of H2O when 0.102 mol Ca(NO3)2 is mixed with 0.927 mol
H2O @ 55.0°C?. PH2O is 118.1 torr
Boiling-Point Elevation and Freezing-Point Depression
The change in boiling point Tb for a solution is
(BPsolution – BPsolvent) = ΔTb = m∙Kb
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The freezing-point depression for a solution relative to that of a pure solvent depends on the
concentration of solute particles, just as boiling-point elevation does.
(FPsolvent – FPsolution) = ΔTf = m∙Kf
ΔTf = Kf • m
For ionic substances: ΔTb = Kb • m • i
ΔTf = Kf • m • i
Example: What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Kf =
1.86 (oC• kg)/mol
FP solution = - 3.2 oC
Example: How many g of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a
solution that boils at 105°C? Kb =0.512 (oC•kg)/mol.
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Example: What is the freezing point in oC of a solution prepared by dissolving 7.40 g of MgCl2 in
110 g of water. Kf = 1.86 (oC• kg)/mol and the van’t Hoff factor for MgCl2 is i=2.7
Example: Determine the freezing in oC for the solution above, assuming MgCl2 have completely
ionized.
Osmosis and Osmotic Pressure
Osmosis: The passage of solvent through a semipermeable membrane from the less concentrated
side to the more concentrated side.
- Solvent flows from a high solvent concentration to a low solvent concentration. Or,
solvent flows from a low solute concentration to a high solute concentration.
- the level of the solution with the higher solute concentration rises.
- the concentrations of the two solutions become equal with time.
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Osmotic Pressure (P): The
amount of pressure necessary to
cause osmosis to stop.
- Osmotic pressure can be
calculated using molarity
since it is calculated at a
specific temperature and
thus no need for
temperature-independent
units.
Π = MRT
R = 0.08206 L•atm/K•mol
Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution.
Starch is a colloid and cannot pass through the membrane, but water can. What happens?
Example: What is the osmotic pressure (in atmospheres) of a 12.36 M aqueous solution of urea at
22.0oC?
Example: What is the molar mass of a protein if 5.87 mg per 10.0 mL gives an osmotic pressure of
2.45 torr at 25°C?
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